Dynamics Example: Force and Acceleration Analysis of a Block and Pulley System

Example 1 Scenario: The problem involves a system of two blocks (A and B) connected by a cord and pulley system, as illustrated in Fig. 13-10a.
Given Masses: * Block A: $100\,\text{kg}$ * Block B: $20\,\text{kg}$
Initial State: The system is released from rest, meaning initial velocity $v_0 = 0\,\text{m/s}$ .
Simplifying Assumptions: * The masses of the pulleys are neglected ( $m = 0$ ). * The mass of the cord is neglected.
Objective: Determine the velocity of the $20\,\text{kg}$ block B at a time interval of $t = 2\,\text{s}$ .

Pulley C Analysis: * Because the mass of the pulleys is neglected, for pulley C, $m_C = 0$ . * This allows the application of the equilibrium equation $\sum F_y = 0$ for the pulley, as shown in Fig. 13-10b. * The analysis shows that if the tension in the cord supporting block B is $T$ , the tension acting on pulley C (and consequently on block A via the two cord segments) involves a balanced force of $2T$ .
Block A FBD (Fig. 13-10c): * Weight: $W_A = 100\,\text{kg} \times 9.81\,\text{m/s}^2 = 981\,\text{N}$ . * Tension: Upward force provided by two segments of the cord, totaling $2T$ .
Block B FBD (Fig. 13-10d): * Weight: $W_B = 20\,\text{kg} \times 9.81\,\text{m/s}^2 = 196.2\,\text{N}$ . * Tension: Upward force provided by a single segment of the cord, $T$ .
Static Equilibrium Check: * For block A to remain stationary, the tension would need to be $T = 490.5\,\text{N}$ (calculated from $2T = 981\,\text{N}$ ). * For block B to remain stationary, the tension would need to be $T = 196.2\,\text{N}$ . * Since these required tensions differ, the system will move. Specifically, block A will move downward while block B moves upward.
Coordinate Assumption: While actual motion direction is known, for the purpose of the derivation, both blocks are assumed to accelerate downward in the direction of $+s_A$ and $+s_B$ .

Analysis for Block A: * Direction of interest: Vertical ( $y$ ) where downward is positive ( $+\downarrow$ ). * Equation: $+\downarrow \sum F_y = m_A a_A$ * Expression: $981 - 2T = 100 a_A$
Analysis for Block B: * Direction of interest: Vertical ( $y$ ) where downward is positive ( $+\downarrow$ ). * Equation: $+\downarrow \sum F_y = m_B a_B$ * Expression: $196.2 - T = 20 a_B$
System Variables: There are three unknowns in these equations: $T$ , $a_A$ , and $a_B$ . A third equation is required to solve the system.

Dependent Motion Definition: The third necessary equation is obtained by relating the acceleration of block A ( $a_A$ ) to the acceleration of block B ( $a_B$ ) using the principles discussed in Sec. 12.9.
Position Coordinates: The coordinates $s_A$ and $s_B$ (see Fig. 13-10a) measure the positions of the blocks relative to the fixed datum.
Total Cord Length ( $l$ ): The total vertical length of the cord is constant and can be expressed by the geometry of the system: * $2s_A + s_B = l$
Time Derivatives: * First derivative with respect to time (Velocity): $2 v_A + v_B = 0$ * Second derivative with respect to time (Acceleration): * $2 a_A + a_B = 0$ * This yields the kinematic constraint: $2 a_A = -a_B$ (Equation 3).

Solving the Linear System: By solving the equations of motion (1 and 2) simultaneously with the kinematic constraint (3), we find the following values: * Tension: $T = 327.0\,\text{N}$ * Acceleration of Block A: $a_A = 3.27\,\text{m/s}^2$ * Acceleration of Block B: $a_B = -6.54\,\text{m/s}^2$
Interpreting the Acceleration: The negative sign for $a_B$ indicates that block B accelerates upward, opposite to the assumed downward "+s" direction.
Final Velocity Calculation for Block B: * Using the constant acceleration kinematic formula: $v = v_0 + a_B t$ * Initial velocity $v_0 = 0$ * Time $t = 2\,\text{s}$ * Calculation: $v = 0 + (-6.54\,\text{m/s}^2)(2\,\text{s})$ * Result: $v = -13.1\,\text{m/s}$ * Conclusion: Block B is traveling at a speed of $13.1\,\text{m/s}$ in the upward direction after $2\,\text{s}$ .